91Ó°ÊÓ

1. Current through the wire,i=2.0A.
2. The radius, r=15cm=0.15m
3. The number of turns,N=6.
4. Uniform external magnetic field,$B=70mT=70×{10}^{-3}T$
5. At exactly 1:00 pm, the hour hand of the clock points in the direction of the external magnetic field.

By using the vector cross product $\mathrm{μ}⃗×B⃗$and applying the right-hand rule, we can find theinterval after which the minute hand will point in the direction of the torque on the winding due to the magnetic field. By using the formula for the magnitude of the torque, we can find the magnitude of the torque.

Formula:

1. The torque is given by$\mathrm{τ}⃗=\mathrm{μ}⃗×B⃗$
2. The magnitude of the torque is given by$\mathrm{τ}⃗=\left|\mathrm{μ}⃗×B⃗\right|$
3. The magnitude of magnetic moment is$\mathrm{μ}=NiA$

Consider the torque is given by:

$\mathrm{τ}⃗=\mathrm{μ}⃗×B⃗$

Since the current goes clockwise around the clock, $\stackrel{→}{\mathrm{μ}}$points to the wall.

Since role="math" localid="1662907647478" $\stackrel{→}{B}$points toward one-hour or 5-minute mark, by the property of the vector product, $\stackrel{→}{\mathrm{τ}}$must be perpendicular to it.

Thus, by using the right-hand rule, we can say that$\mathrm{Ï„}$points at the 20-minute mark.

So, the time interval is 20min.

The magnitude of the torque is given by

$$\begin{gathered} \mathrm{τ}=|\mathrm{μ}⃗×B⃗| \\ \mathrm{τ}=\mathrm{μ}Bsin{90}^0 \\ \mathrm{τ}=\mathrm{μ}B \end{gathered}$$

…… (1)

But, the magnitude of magnetic moment is

$\mathrm{μ}=NiA$

With$A=\mathrm{Ï€}{r}^2$,

$\mathrm{μ}=\mathrm{π}Ni{r}^2$

Substitute the values in equation

$\mathrm{Ï„}=\mathrm{Ï€}Ni{r}^{2}B$

Substitute the values and solve as:

$\mathrm{τ}=\mathrm{π}×6×(2.0 A)×(0.15 m{)}^2×(70×{10}^{-3}T)$

$\mathrm{Ï„}=5.9×{10}^{-2}Nâ‹\dots m$